For $R$-modules $M,N$, the <tensor product of modules> is an $R$-module $M\otimes_RN$ together with the balanced map
$$
\tau:M\times N\longrightarrow M\otimes_RN,
\qquad
(m,n)\longmapsto m\otimes n,
$$
such that every <balanced map> $b:M\times N\to P$ factors through one unique $R$-linear map $\widetilde b$:
$$
b=\widetilde b\circ\tau.
$$
Equivalently,
$$
\operatorname{Hom}_R(M\otimes_RN,P)
\cong\{\text{balanced maps }M\times N\to P\}
$$
naturally in $P$. This is the <universal property of the tensor product of modules>.

Solved by gpt-5.6-sol high.

Solution

Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 101 / 1 / i / Solution

The <Tensor-Hom adjunction> is the natural isomorphism
$$
\Phi_{M,N,P}:\operatorname{Hom}_R(M\otimes_RN,P)
\longrightarrow
\operatorname{Hom}_R\bigl(M,\operatorname{Hom}_R(N,P)\bigr).
$$
For an <R-module homomorphism> $f$, it is given explicitly by
$$
\Phi(f)(m)(n)=f(m\otimes n).
$$
Conversely, an $R$-linear map $g:M\to\operatorname{Hom}_R(N,P)$ determines the <balanced map> $(m,n)\mapsto g(m)(n)$, so the <universal property of the tensor product of modules> gives
$$
\Phi^{-1}(g)(m\otimes n)=g(m)(n).
$$
These formulas are inverse to each other because pure tensors generate the <tensor product of modules>.

Solved by gpt-5.6-sol high.

Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 101 / 1 / a / Solution

{title2=$V^{\otimes n}$}
{wiki=Tensor_product#Tensor_powers_and_braiding}

The nth tensor power is the iterated <tensor product of modules> $V^{\otimes n}=V\otimes\cdots\otimes V$.

Tensor power

{title2=$M\otimes_RN$}
{wiki=Tensor_product}

The tensor product of an $R$-module $M$ and an $R$-module $N$ is an $R$-module $M\otimes_RN$ generated by pure tensors $m\otimes n$, subject to additivity in both variables and the balancing relation
$$
(rm)\otimes n=m\otimes(rn).
$$


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 Incoming links: Tensor product of modules